Question: $\text F = \left[\begin{array}{rr}1 & 2 \\ -2 & 3\end{array}\right]$ and $\text E = \left[\begin{array}{rrr}0 & -1 & 5 \\ 3 & 2 & 1\end{array}\right]$ Let $\text {H = FE}$. Find $\text H$. $ {H = }$
Solution: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{F}$ and the first column of $\text{E}$. $ \text {H}=\left[\begin{array}{rr}{1} & {2} \\ -2 & 3\end{array}\right]\left[\begin{array}{rr} {0} & -1 & 5 \\ {3} & 2 & 1\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(1,2)\cdot(0,3)\\\\ &=1 \cdot 0 + 2\cdot 3\\\\ &=6 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-2 \cdot -1 + 3\cdot 2 = 8$ (Choice B) B $1 \cdot -1 + 2\cdot 2 = 3$ (Choice C) C $-2 \cdot 0 + 3\cdot 3 = 9$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}= \left[\begin{array}{rrr}6 & 3 & 7 \\ 9 & 8 & -7\end{array}\right] $